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Calculus Basics 微积分基础 #49

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solomonxie opened this issue May 11, 2018 · 95 comments
Open
20 of 26 tasks

Calculus Basics 微积分基础 #49

solomonxie opened this issue May 11, 2018 · 95 comments

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@solomonxie
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solomonxie commented May 11, 2018

Notes on basic Calculus concepts.

Study resources

Study Tools

Practice To-do List (Linked with Unit tests)

Table of Contents

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solomonxie commented May 11, 2018

「Limits」 properties

Refer to Khan academy: Limit properties

image

The limit of a SUM of functions is the SUM of the INDIVIDUAL limits:
image

Limits of 「Combined Functions」

Refer to Khan academy: Limits of combined functions

Example

image
Solve:
image

Example

image
Solve:

  • The limit for each function DOES NOT exists.
  • However, the one-side limits of each graph DO exist
    image
  • We can use the fact that the limit of a sum of functions is the sum of the individual limits:
    image
  • Calculate each side's combined limits:
    image

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solomonxie commented May 12, 2018

Function's Continuity [DRAFT]

Pencil Definition:
IF YOU CAN DRAW THE FUNCTION WITH A PENCIL WITHOUT PICKING UP THE PENCIL, THEN THE FUNCTION IS A CONTINUOUS FUNCTION

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solomonxie commented May 15, 2018

Limits

Limits are all about approaching.
And the entire Calculus is built upon this concept.

Strategy in finding Limits

Refer to Khan academy.

image

@solomonxie
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solomonxie commented May 15, 2018

Limits at 「infinity」

No matter why kinds of Limits you're looking for,
to understand it better,
the best way is to read the Step-by-Step Solution from Symbolab:
Limit Calculator from Symbolab.

「Rational functions」

The KEY point is to look at the powers & coefficients of Numerator & Dominator.
Just the same with Finding the Asymptote.

Refer to previous note on the How to find Asymptote.

image

Example

image
Solve:

Quotients with 「square roots」

The KEY point is to calculate both numerator & dominator, then calculate the limit of EACH term with in the square root.

Example

image
Solve:
Refer to Symbolab step-by-step solution.

  • Divide by highest dominator power to get:
    image
  • Calculate separately the limit of Numerator & Dominator:
    image
  • Calculate the Square root: Need to find limits for EACH term inside the square root.
    image
    image
    image
  • Then get the result easily.

Quotients with 「trig」

The KEY point is to apply the Squeeze theorem, and it is a MUST.

Example

image
Solve:

  • Know that -1 ≦ cos(x) ≦ 1, so we can tweak it to apply the squeeze theorem to get its limit.
  • Make the inequality to: 3/-1 ≦ 3/cos(x)/-1 ≦ 3/1
  • Get that right side 3/-1 = -1 and left side 3/1 =1 is not equal.
  • So the limit doesn't exist.

Easier solution steps:

  • Know the inequality -1 ≦ cos(x) ≦ 1
  • Replace cos(x) to ±1 in the equation, 3/±1.
  • Calculate limits of two sides.
  • If the results are exactly the same, then the limit is the result; Otherwise the limit doesn't exist.

Example

image
Solve:

  • Know that -1 ≦ sin(x) ≦ 1
  • Replace sin(x) as ±1
  • Left side becomes (5x+1)/(x-5), right side becomes (5x-1)/(x-5)
  • Both sides' limits are 5, so the limit exists, and is 5.

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solomonxie commented May 15, 2018

❖ All types of discontinuities

Refer to Mathwarehouse: What are the types of Discontinuities?

「Jump Discontinuities」

A Jump discontinuity occurs at some point, the left side limit is DIFFERNT with the right side limit.

image
We see that clearly:
image

「Removable Discontinuities」

A removable discontinuity occurs at some point, both left side limit & right side limit are the SAME.

image

「Infinite Discontinuities」

An Infinite Discontinuity occurs at some point, both left side & right side are approaching to INFINITY, which means both sides DO NOT have limits.

image

「Endpoint Discontinuities」

An Endpoint Discontinuity occurs at some point, it DOESN'T HAVE both sides, it only has ONE SIDE.

image

「Mixed Discontinuities」

image

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solomonxie commented May 15, 2018

Analyzing functions for discontinuities「algebraic」

If the limits of both side of some point, are EQUAL, then it's continuous at this point.

At a point a, for f(x) to be continuous at x=a, we need lim(x→a)f(x) = f(a).

Example

image
Solve:

  • Got the limit of left side is -1/2
  • Got the limit of right side is -1/2
  • They're equal, so it's continuous at this point.

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solomonxie commented May 15, 2018

Application of Removable Discontinuities

Example

image
Solve:
It's just the same with calculating the limits of both sides.

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solomonxie commented May 16, 2018

❖ 「Derivative」 Basics

Simply saying, it's just the SLOPE of ONE POINT of a graph (line or curves or anything).

Refer to Mathsisfun: Introduction to Derivatives

image

A Derivative, is the Instantaneous Rate of Change, which's related to the tangent line of a point, instead of a secant line to calculate the Average rate of change.

“Derivatives are the result of performing a differentiation process upon a function or an expression. ”

Derivative notations

Refer to Khan academy article: Derivative notation review.

image

「Lagrange's」 notation

In Lagrange's notation, the derivative of f(x) expressed as f'(x), reads as f prime of x.

「Leibniz's」 notation

In this form, we write dx instead of Δx heads towards 0.
And the derivative of is commonly written as:
image

For memorizing, just see d as Δ, reads Delta, means change. So dy/dx means Δy/Δx. Or it can be represent as df / dx or d/dx · f(x), whatever.

image

image

How to understand 「dy/dx」

Refer to Khan academy from Differential Equation section: Addressing treating differentials algebraically

This is a review from "the future", which means while studying Calculus, you have to come back constantly to review what the dy/dx means. ---- It's just so confusing.
Without fully understanding the dy/dx, you will be lost at topics like Differentiate Implicit functions, Related Rates, Differential Equations and such.

image

「Tangent line」 & 「Secant line」

  • The secant line is drawn to connect TWO POINTS, and gets us the Average Rate of Change between two points.
  • The Tangent line is drawn through ONE POINT, and gets us the Rage of change at the exact moment.

As for the secant line, its interval gets smaller and smaller and APPROACHING to 0 distance, it actually is a process of calculating limits approaching 0, which will get us the tangent line, that been said, is the whole business we're talking about: the Derivative, the Instantaneous Rate of Change.

image

image

secant to tangent animation

「Secant line」

Example

image
Solve:

  • What it's asking is the Slope of its secant line:
    image
  • which could be applied with this simple formula:
    image
  • the result is ( f(3)-f(1) )/ (3-1) = -1/12

@solomonxie solomonxie added the HF High frequency notes label May 16, 2018
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solomonxie commented May 16, 2018

❖ Differentiability

"If the point of a function IS differentiable, then it MUST BE continuous at the point."

Example of NOT differentiable points:
image

You can see, if the point DOES NOT have limit, it's NOT DIFFERENTIABLE.
In another word, the point is not CONTINUOUS, it's Jump Discontinuity, or Removable Discontinuity, or any type of discontinuities.

「NOT」 differentiable situations

  • Vertical Tangent (∞)
  • Not Continuous
  • Two sides' limits are different

image

「Vertical Tangent」

We know that the Slope of Vertical Tangent is UNDEFINED,
on the contrary:
IT IS A VERTICAL TANGENT, IF:

  • The derivative dy/dx = undefined, or
  • The denominator of derivative's expression = 0.

「Horizontal Tangent」

It's a Horizontal Tangent, if:

  • dy/dx = 0.

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solomonxie commented May 17, 2018

❖ Derivative equation

The idea of derivative equation is quite simple: The LIMIT of the SLOPE.

The slope is equal to change in Y / change in X.
So for a point a, we IMAGINE we have another near point which lies on the SAME LINE with a,
and since we have TWO POINTS now,
we can then let their Y-value Change divided by their X-value Change to get the slope.

There're two equations for calculating derivative at a point, and the only different thing is how to express the IMAGINARY POINT with respect to the point a, it could either be x or a+h :

image
or:
image

How to calculate 「derivative」

Strategy:

  • Determine if it's CONTINUOUS at this point, by:
    • See if the point is defined in the interval
    • Calculate LIMITS of both RIGHT SIDE and LEFT SIDE of the point.
    • If two sides' limits are the same, then it's continuous. Otherwise it's discontinuous.
  • Determine if it's DIFFERENTIABLE (Actually is the process of getting its derivative):
    • Apply Derivative equation to get both RIGHT SIDE LIMIT and LEFT SIDE LIMIT.
    • If two sides' limits are the same, then that value is the Derivative at the point. If not, then it's NOT DIFFERENTIABLE.

Example

image
Solve:

  • See that the point 3 is defined in the interval.
  • Left side limit of the point, is using the first equation, and gets the lim g(x) = -7
  • Right side limit of the point, is using the second equation, and gets the lim g(x) = -7
  • Limits of both sides are the SAME, so it's continuous, and let's see if it's differentiable.
  • Apply the derivative equation for both Left side & Right side:
    image
  • Both sides' limits exists but not that same, so it's not differentiable.

Example

image
Solve:

  • See that the point -1 is defined in the interval.
  • Left side limit of the point, is using the first equation, and gets the lim g(x) = 1
  • Right side limit of the point, is using the second equation, and gets the lim g(x) = 4
  • Limits of both sides are NOT SAME, so it's not continuous, then of course not differentiable.

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solomonxie commented May 17, 2018

「Local linearity」 & 「Linear approximation」

Local linearity is for approximating of a point's value by its near known point.

Just think of a curve, a good way to approximate its Y-value, is to find another known point near it, and make a line connecting two points, then gets the value by linear equation.

Refer to Khan academy lecture.

image

image

Example

image
Solve:
image

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solomonxie commented May 18, 2018

Basic 「Differential Rules」

image

Jump over to Basic Integral Rules

▼Refer to Math is fun: Derivative Rules

Rules Function Derivative
Constant c 0
With constant c·f c·f’
Power Rule xⁿ n·xⁿ⁻¹
Sum Rule f + g f’ + g’
Difference Rule f - g f’ − g’
Product Rule f · g fg’ + f’ g
Quotient Rule f / g (f’g − g’f ) / g²
Reciprocal Rule 1 / f −f’ / f²
Chain Rule f(g(x)) f’(g(x)) · g’(x)
Exponent Rule
                      aˣ · ln(a)
Log Rule image image
Natural Log Rule ln(x) 1/x
Exponential Rule image image
Trig Rules sin(x) cos(x)
                     cos(x) −sin(x)
                      tan(x) sec²(x) = 1/cos²(x)
                      sec(x) sec(x)·tan(x)
                      csc(x) -csc(x)·cot(x)
                    cot(x) −csc²(x) = -1/sin²(x) = -(1+cot²(x))
Inverse Trig Rules arcsin(x) 1/√(1−x²)
                               arccos(x) −1/√(1−x²)
                                arctan(x) 1/(1+x²)

▼Refer to Wiki: Differentiation rules
image

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solomonxie commented May 18, 2018

Chain Rule

One of the core principles in Calculus is the Chain Rule.

image

Refer to Khan academy article: Chain rule
▶ Proceed to Integral rule of composite functions: U-substitution

It tells us how to differentiate Composite functions.

image

It must be composite functions, and it has to have inner & outer functions, which you could write in form of f(g(x)).
image

Common 「mistakes」

  • Not recognizing whether a function is composite or not
  • Wrong identification of the inner and outer function
  • Forgetting to multiply by the derivative of the inner function
  • Computing f(g(x)) wrongly:
    image

How to identify 「Composite functions」

Seems a basic algebra101, but actually a quite tricky one to identify.

Refer to Khan lecture: Identifying composite functions

The core principle to identify it, is trying to re-write the function into a nested one: f(g(x)). If you could do this, it's composite, if not, then it's not one.

Examples

image
It's a composite function, which the inner is cos(x) and outer is .

image
It's a composite function, which the inner is 2x³-4x and outer is sin(x).

image
It's a composite function, which the inner is cos(x) and outer is √(x).

Two forms of 「Chain Rule」

The general form of Chain Rule is like this:
image

But the Chain Rule has another more commonly used form:
image

Their results are exactly the same.
It's just some people find the first form makes sense, some more people find the second one does.

Example

image
Solve:
Refer to Symbolab worked example.

Chain rule for 「exponential function」

Formula:
image

Because:
image

Example

image
Solve:

  • Apply the Log power rule to simplify the exponential function:
    image
  • Differentiate both sides:
    image

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solomonxie commented May 18, 2018

Derivatives of 「Trig functions」

image

Reminder of 「Trig identities」 & 「Unit circle」 values

# Reciprocal and quotient identities
tan(θ) = sin(θ) / cos(θ)
csc(θ) = 1/sin(θ)
sec(θ) = 1/cos(θ)
cot(θ) = 1/tan(θ)

Refer to previous note of all trig identities.

unit circle

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solomonxie commented May 18, 2018

❖ Implicit Differentiation

Bit hard to understand it in the first place.

image

What is 「Implicit」 & 「Explicit Function」

Refer to video by Krista King: What is implicit differentiation?

  • Explicit function: it's the normal function we've seen a lot before, which's in the form of y = x....
  • Implicit function: it't NOT YET in the general form of a function and not easily separated, like x² + y² = 1

So knowing how to differentiate an implicit function is quite helpful when we're dealing with those NOT EASILY SEPARATED functions.

How to Differentiate 「Implicit function」

Refer to video: Use implicit differentiation to find the second derivative of y (y'') (KristaKingMath)
Refer to video by The Organic Chemistry Tutor: Implicit Differentiation Explained - Product Rule, Quotient & Chain Rule - Calculus

Refer to Symbolab: Implicit Derivative Calculator

Assume you are to differentiate Y WITH RESPECT to X, written as dy/dx:

  • Differentiate terms with X as normal
  • Differentiate terms with Y as the same to X, BUT multiply by (dy/dx)
  • Differentiate terms MIXED with X & Y by using Product Rule, then differentiate each term.

How to differentiate 「Y with respect to X」

image

How to differentiate 「term MIXED with both X & Y」

image

Example

image
Solve:
Refer to Symbolab: Implicit Derivative Calculator

  • Treat y as y(x)
  • Apply the Sum Rule:
    image
  • Apply the normal rules to X term, and
  • Apply the Product Rule to the Mixed term, and
  • Apply the Chain Rule to the Y term:
    image
  • Operate the equation and solve for dy/dx, and get:
    image

Example

image
Solve:

  • First thing we need to find the RIGHT equation of Chain rule. Since it's asking us to find dy/dt, so we will re-write it to this one to form an equation:
    image
  • Then since we've given the dx/dt = -3, we only need to find out the dy/dx to get the result.
  • We've got an equation of x & y, regardless whom it's respecting to. So we can do either Implicit or Explicit differentiation to the equation y²=7x+1, with respect to y:
    image
  • Use the implicit differentiation method, we got the dy/dx = 7/2y
  • And since y=6, so 7/2y = 7/12
  • Back to the Chain Rule equation, we get dy/dt = 7/12 · (-3) = -7/4 = -1.75

Example

image
Solve:

  • Remind you that, in this problem, it's NOT respecting to x anymore, so you need to change mind before getting confused.
  • First thing we need to find the RIGHT equation of Chain rule. Since it's asking us to find dx/dt, so we will re-write it to this one to form an equation:
    image
  • Then since we've given the dy/dt = -0.5, we only need to find out the dx/dy to get the result.
  • We've got an equation of x & y, regardless whom it's respecting to. It seems easier to differentiate explicitly:
    image
  • Then we use d/dx to differentiate the equation to get: dx/dy = y⁻² = (0.2)⁻² = 25
  • Back to the Chain Rule equation, we get dx/dt = dx/dy · dy/dt = 25 * (-0.5) = -12.5.

Example

image
Solve (Same with above examples):

  • Form an equation:
    image
  • dx/dt has been given equals to 5, so just to find out dy/dx:
    image
  • And get:
    image
    image
  • Now let's see what is sin(x) equal to:
    image
  • All done.

「Vertical & Horizontal Tangents」 of 「Implicit Equations」

► Jump over to Khan academy for practice.

Example

image
Solve:

  • Plug in y = 0 into the equation and get that x = -6, which is the answer.

Example

image
Solve:

  • To have a Vertical Tangent, we have to let the derivative become Undefined,
  • which in this case is to let the denominator equal to zero:
    image
  • Solve this equation out we get that x = 3y², which means this relationship is true at the point of vertical tangent line.
  • Plug that back to the original function to get y = -1, which means the vertical tangent goes through this point.
  • Substitute y back and get x = 3
  • The answer is (3, -1).

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solomonxie commented May 19, 2018

❖ Related Rates

Just so you know, related rates is actually the Application of Implicit Differentiation by using Chain Rule in the form of dy/dx = dy/du * du/dx.

image

Btw, at Khan academy it's called the Differentiate related functions.

Refer to Khan lecture.
Refer to video by KristaKingMath: Related rates
Refer to video by The Organic Chemistry Tutor: Introduction to Related Rates

Strategy:

  • Abstract every given condition algebraically, e.g. s(t), s'(t), V(t), V'(t)...
  • Find a proper equation to CONNECT all these given conditions
  • Differentiate both sides of equation, with using the Chain rule.
  • Take back the given values to the Differentiated equation to get the result.

Example: 「Change of volumes」

Refer to previous note of Implicit Differentiation.
image

Solve:

  • Write out all conditions algebraically:
    • r'(t)=1
    • r(t)=5
    • h'(t)=-4
    • h(t)=8
    • V(t) = π·r²·h
  • So they're asking for V'(t), it then becomes V'(t) = π·d/dt (r²·h)
  • Apply Product rule & Chain rule then substitute: V'(t) = π((r²)'·h + r²h') = π(2r'·r + r²·h') = -20π
    Notice that: (r²)' is a composite function, so you want to apply the Chain rule, =2r·r'

Example: 「Change of volumes」

image
Solve:

  • From the given conditions, we got that r'(t)=-12, r(t)=40 and h=2.5.
  • We know the equation of cylinder's volume is: V= π·[r(t)]²·h
  • Differentiate both side of the equation to get:
    image
  • Take back all the known values into the equation get V'(t)=-2400π

Example: Change of area

image
Solve:

  • Write out all conditions algebraically:
    • d₁'(t) = -7
    • d₁(t) = 4
    • d₂'(t) = 10
    • d₂(t) = 6
    • A(t) = (d₁·d₂)/2
  • So it's asking for instantaneous rate of change, then it is A'(t) = (d₁'·d₂ + d₁·d₂')/2
  • Apply the Product rule only, to get A'(t) = -1.

Example: 「Pythagorean Theorem」

image
image
Refer to Khan academy lecture: Related rates: Approaching cars

Solve:

  • It's a problem to calculate two objects' dynamic absolute distance: it's getting closer and closer until distance become 0, which means they crashes at 0.
  • Write down all the conditions algebraically:
    • c₁'(t) = -10
    • c₁(t) = 4
    • c₂'(t) = -6
    • c₂(t) = 3
    • D(t) = √(c₁²+c₂²)
  • So D'(t) = d/dt · (√(c₁²+c₂²)) = (2c₁'c₁+2c₂'c₂)/(2√(c₁²+c₂²))

Example: 「Sliding ladder」

image

Example: 「multiple composite」

image
Solve:

  • Write down all the conditions algebraically:
    • S'(t) = π/2
    • S(t) = 12π
    • S = 2π·r, which means r = S/2π
    • A(t) = π·r²
  • So A'(t) = d/dt · (π·r²) = 2π·r·r' = 2π·6·(S/2π)' = 6S' = 3π

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solomonxie commented May 19, 2018

❖ Higher Order Derivatives

  • First order derivative: f'(x) or dy/dx, also called "First Derivative"
  • Second order derivative: f''(x) or d²y/dx², also called "Second Derivative"
  • ...

「Second derivatives」

The second derivative of a function is simply the derivative of the function's derivative.

Notation:
Leibniz's notation for second derivative is:
image
image

Second derivatives of 「implicit equations」

▶ Jump back to previous note on Implicit Differentiation
▶ Practice at Khan academy: Second derivatives (implicit equations)

▶ Online calculator for 2nd Derivative of Implicit equations

Example

image
Solve:
image

Example

image
Solve:

  • The first derivative is:
    image
  • The second derivative is:
    image
  • Plug in y=1, and get:
    image

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solomonxie commented May 21, 2018

Derivative of Inverse functions

image

IT'S DERIVED FROM THE CHAIN RULE:

image

Derivative of 「Inverse Trig functions」

image

Example

image
Solve:

  • For solving this derivative of an inverse function, we'd use the identity:
    image
  • Apply it we get:
    image

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solomonxie commented May 21, 2018

Derivative of 「exponential functions」

It's save a lot of time of life not to dig in how mathematicians developed these formulas.
If you do want to, refer to Khan's lecture: Exponential functions differentiation intro

Reminder: Don't forget it's a composite function and you need to apply the chain rule.

image

Derivative of 「log functions」

image

Examples

Find the derivative of:
image
Solve:

image

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solomonxie commented May 22, 2018

❖ Existence Theorems (IVT, EVT, MVT)

Existence theorems includes 3 theorems: Intermediate Value Theorem, Extreme Value Theorem, Mean Value Theorem.

Refer to Khan academy: Existence theorems intro

image

「Intermediate Value Theorem」 (IVT)

The IVT is saying:

When we have 2 points connected by a continuous curve:
one point below the line, the other point above the line,
then there will be at least one place where the curve crosses the line!

image

Refer to Maths if fun: Intermediate Value Theorem
Refer to video: Intermediate Value Theorem Explained

Find 「roots」 by using IVT

IVT is often to find roots of a function, which means to find the x value when f(x)=0.
So for finding a root, the definition will be:

If f(x) is continuous and has an interval [a, b], which leads the function that f(a)<0 & f(b)>0 , then it MUST has a point f(c)=0 between interval [a,b], which makes a root c.

Example

Tell whether the function f(x) = x² - x - 12 in interval [3,5] has a root.
Solve:

  • We got that at both sides of intervals: f(3)=-6 < 0, and f(5)=8 > 0
  • So according to the Intermediate Value Theorem, there IS a root between [3,5].
  • Calculate f(c)=0 get the root c=4.

「Extreme Value Theorem」 (EVT)

The EVT is saying:

There MUST BE a Max & Min value,
if the function is continuous over the closed interval.

image

Refer to Khan lecture: Extreme value theorem
Refer to video: Extreme Value Theorem

「Mean Value Theorem」 (MVT)

Refer to Khan academy article: Establishing differentiability for MVT

The MVT is saying:

There MUST BE a tangent line that has the same slope with the Secant line,
if the function is CONTINUOUS over [a,b] and DIFFERENTIABLE over (a,b).

Which also means that, if the conditions are satisfied, then there MUST BE a number c makes the derivative is equal to the Average Rate of Change between the two end points.

Equation

Conditions for applying MVT:

  • Continuous over interval (a, b)
  • Differentiable over interval [a, b]

Graph

Example

image
Solve:

  • He's totally right.

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solomonxie commented May 23, 2018

❖ L'Hopital's Rule [DRAFT]

LHopital's Rule helps us to find the limit of an Undefined limits, like 0/0, ∞/∞ and such.
It's quite simple to apply and very convenient to solve some problems.

Refer to L'Hôpital's rule

▶ Jump back previous note on: Asymptote of Rational Expressions
▶ Practice at Khan academy: Disguised derivatives

From my experience, the L'Hopital's Rule is so often been used that we didn't even realize. Actually it's been used almost every time when we are to evaluate the LIMITS OF RATIONAL EXPRESSIONS.

image

Example of 「0/0」

Example of 「∞/∞」

  1. Find the limit:
    image
    Solve:
    image

  2. Find the limit:
    image
    Solve:
    image

Example of 「1^∞」

L'Hopital Rule for 「Composite functions」

Example of composite exponential function

image
Solve:

  • Direct plug in the x=2 and get limit = 1^∞, which is an indeterminate form
  • So we're gonna apply the Log Power Rule to take down the power:
    image
  • We use Natural Log for doing this:
    image
  • And now we can apply the L'Hopital Rule for ln(y):
    image
  • From ln(y) we could get limit of y:
    image

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solomonxie commented May 23, 2018

❖ Critical points

Refer to PennCalc Main/Optimization

For analyzing a function, it's very efficient to have a look at its Critical points, which could be classified as Extrema, Inflection, Corner, and Discontinuity.

image

How to find 「critical points」

Strategy:

  • Knowing that f(x) has critical point c when f'(c) = 0 or f'(c) is undefined
  • Differentiate f(x) to get f'(x)
  • Solve c for f'(c)=0 & f'(c) undefined

Refer to Symbolab's step-by-step solution.

Example

image
Solve:

  • See that original function f(x) is undefined at x = 2 or -2
  • Differentiate f(x) to get f'(x):
    image
  • Solve f'(x)=0 only when x=0.
  • f'(x) is undefined when x=2 or -2, as the same with f(x) so it's not a solution.

Example

image
Solve: Refer to Symbolab step-by-step solution.

  • Differentiate f(x) to get f'(x):
    image
  • f'(x) is undefined when x > 4
  • Solve f'(x)=0 get x = 8/3
  • So under the given condition, only x=8/3 is the answer.

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solomonxie commented May 25, 2018

❖ Extrema: 「Maxima」 & 「Minima」

Extrema are one type of Critical points, which includes Maxima & Minima.
And there're two types of Max and Min, Global Max & Local Max, Global Min & Local Min.
We can all them Global Extrema or Local Extrema.

And actually we can call them in different ways, e.g.:

  • Global Max & Local Max or in short of glo max & loc max
  • Absolute Max & Relative Max or in short ofabs max & rel max

image

How to identify 「Extrema」

We need two kind of conditions to identify the Max or Min.
Now If we have a Non-Endpoint Minimum or Maximum point at a, then it must satisfies these conditions:

  • Geometric condition: (It should be understood in a more intuitive way)
    • in the interval [a-h, a+h] there's no point above or below f(a) or
    • f'(a-h) & f'(a+h) have different sign, one negative another positive.
  • Derivative condition:
    • f'(a) = 0 or
    • f'(a) is undefined

How to find 「Extrema」

Refer to Khan academy lecture: Finding critical points

We just need to assume f'(x) = 0 or f'(x) is undefined, and solve the equation to see what x value makes it then.

「Increasing」 & 「Decreasing Intervals」

We can easily tell at a point of a function, it's at the trending of increasing or decreasing, by just looking at the instantaneous slope of the point, aka. the derivate.
If the derivative, the slope is positive, then it's increasing. Otherwise it's decreasing.

Finding the 「trending」 at a point

Just been said above, we assume at point a, it's value is f(a). So the slope of it is f'(a).
And if f'(a) < 0, then it's decreasing; If f'(a) > 0, then it's increasing.

Finding a 「decreasing or increasing interval」

Refer to Khan lecture.

It's just doing the same thing in the opposite way.
For find a decreasing interval, we assume f'(x) < 0, and by solving the inequality equation we will get the interval.

Strategy:

  • Get Critical points.
  • Separate intervals according to critical points, undefined points and endpoints.
  • Try easy numbers in EACH intervals, to decide its TRENDING (going up/down):
    • If f'(x) > 0 then the trending of this interval is Increasing.
    • If f'(x) < 0 then the trending of this interval is Decreasing.

Example

image
Solve:

  • Set f'(x) = 0 or undefined, get x = -2 or -1/3
  • Separate intervals to (-∞, -2, -1/3, ∞)
    image
  • Try some easy numbers in each interval: -3, -1, 0:
    image

How to find 「Relative Extrema」

Remember that an Absolute extreme is also a Relative extreme.

Refer to khan: Worked example: finding relative extrema
Refer to Khan Academy article: Finding relative extrema

image

Strategy:

  • Get Critical points.
  • Separate intervals according to critical points, undefined points and endpoints.
  • Try easy numbers in EACH intervals, to decide its TRENDING (going up/down).
  • Decide each critical point is Max, Min or Not Extreme.

Refer to an awesome article: Using calculus to learn more about the shapes of functions

image

Example

image
Solve:

  • Set f'(x)=0 or undefined, get x=0 or -2 or 1
  • Separate intervals to (-∞, -2, 0, 1, ∞)
    image
  • Try easy number in each interval: -3, -1, 0.5, 2 and get the trendings:
    image
  • Identify critical points' concavity:
    image

How to find 「Absolute Extrema」

Refer to Khan academy article: Absolute minima & maxima review
Refer to Khan academy lecture: Finding absolute extrema on a closed interval

Strategy:

  • Find all Relative extrema
    • Get Critical points.
    • Separate intervals according to critical points & endpoints.
    • Try easy numbers in EACH intervals, to decide its TRENDING (going up/down).
    • Decide each critical point is Max, Min or Not Extreme.
  • Input all the extreme point into original function f(x) and get extreme value.

Example

image
Solve:

  • Set g'(x) = 0 or undefined get x=0
  • Separate intervals to [-2, 0, 3] according to the critical point & endpoints of the given condition.
  • Try some easy numbers of each interval -1, 2 into g'(x)
  • Get the trending of each interval: (+, +)
  • So the minimum must be the left endpoint of the given condition, which is x=-2.

Example

image
Solve:

  • Differentiate to set f'(x)=0 and got x=0, -1, 1
  • Separate intervals to (-∞, -1, 0, 1, +∞)
  • Try some easy numbers in each interval for f'(x) and got the signs: -, +, -, +
  • According to the signs we know that -, + gives us a Relative maximum
  • But at the end it's + again, and the interval is going up to +∞, means f(x) will go infinitely high.
  • So there's NO Absolute maximum.

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solomonxie commented May 25, 2018

Concavity

Refer to Khan academy: Concavity introduction

Two types of concavity: Concave Up & Concave Down.

Understand function by 「1st & 2nd Derivative」

image

Identify 「concavity」 by using 「Second Derivative」

image

Example

image
Solve:

  • f'(x) > 0 means we're to find the INCREASING interval on the graph
  • f''(x) < 0 means we'll be focusing on the CONCAVE UP shape only
  • So it's about the interval of(-4, -3) and (3, 4).

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solomonxie commented May 25, 2018

❖ Inflection Point

An inflection point is a point where the graph of the function changes CONCAVITY (from up to down or vice versa).

It could be seen as a Switching point, which means the point that the Slope of function switch from increasing and decreasing.
e.g., the function might be still going up, but at such a point it suddenly increases slower and slower. And we call that point an inflection point.
image

Refer to Khan academy video for more intuition rapidly: Inflection points from graphs of function & derivatives

Algebraically, we identify and express this point by the function's First Derivative OR Second Derivative.
image

Example

image
Intuitive way to solve:

  • Draw a tangent line in imagination and move it on the function from left to right
  • Notice the tangent line's slope, does it go faster or slower or suddenly change its pace at a point?
  • We found it suddenly changed at point c.

More definitional way to solve:

  • Looking for the parts of concavity shapes
  • Seems that B-C is a part of Concave Down, and C-D is a part of Concave Up
  • So C is a SWITCHING POINT, it's a inflection point.

Example

image
Solve:

  • Looking for the parts of concavity shapes
  • There's no changing of concavity shapes, there's only one shape: Concave down.

Example

image
Solve:

  • Notice that's the graph of f'(x), which is the First Derivative.
  • Checking Inflection point from 1st Derivative is easy: just to look at the change of direction.
  • Obviously there're only two points changed direction: -1 & 2

Example

image
Solve:

  • Mind that this is the graph of f''(x), which is the Second derivative.
  • Checking inflection points from 2nd derivative is even easier: just to look at when it changes its sign, or say crosses the X-axis.
  • Obviously, it crosses the X-axis 5 times. So there're 5 inflection points of f(x).

Example: Finding Inflection points

image
Solve:

  • Function has POSSIBLE inflection points when f''(x) = 0.
  • Set f''(x) =0 and solve for x, got x=-3.
  • We now know the possible point, but don't know its CONCAVITY. This need to try some numbers from its both sides:
    image
  • So it didn't change the concavity at point -3, means there's no inflection point for function.

Example: Finding Inflection points

image
Solve:

  • Function has POSSIBLE inflection points when f''(x) = 0.
  • Set f''(x) =0 and solve for x, got x=0 or 6.
    Refer to Symbolab for f''(x).
    Refer to Symbolab for f''(x)=0.
  • We now know the possible point, but don't know its CONCAVITY. This need to try some numbers from its both sides:
    image
  • So it didn't change the concavity at point 0, means only 6 is the inflection point.

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solomonxie commented Jun 23, 2018

❖ Infinite Seires

image

Evaluate 「Infinite Series」 as limit of 「partial sums」

Evaluate the series, is actually to evaluate the LIMIT of the series function.

Example

image
Solve:
image

Example

image
Solve:
image

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solomonxie commented Jun 23, 2018

Infinite Geometric Series

Common Formula (Finite & Infinite)

image

Infintite Formula

image

Determine the Infinite Geometric Series Converges or Diverges

There are two basic rules for infinite geometric series:
image

Example

image
Solve:

  • Yes. Because |-0.8| < 1, satisfies the basic converge rules.

Evaluate Infinite Geometric Series

Example

image
Solve:

  • First off, we need to find the essential information for the geometric series:
    • Initial term: a₀ = 1
    • Common Ratio: r = 0.75
  • So the Infinite series would be:
    image

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solomonxie commented Jun 23, 2018

❖ Convergence Tests [DRAFT]

Here lists common Convergence Tests and overview of each. Details are singled out to each section.

Convergence test are a set of tests to determine wether the series CONVERGENT or DIVERGENT.
It includes:

Test Description
► Divergent Test Take nth term's limit. (only to test divergence)
► Integral Test Take limit of the series function's integration.
► p-series Test Examine at the p value of 1/nᴾ.
► Comparison Test Compare the series to a "similar" p-series or geometric-series.
► Ratio Test Take limit of two terms ratio.
► Root Test Take the limit of nth root of nth term.
► Alternating Test Test if terms are decreasing, and take limit of nth term.

「Divergent Test」

Take the limit of nth term, if it's NOT ZERO, then it's DIVERGENT.

image

「Integral Test」

「p-series Test」

image

「Direct Comparison Test」

Compare the series to a "similar" p-series or geometric-series.

image

「Limit Comparison Test」

Compare the series to a "similar" p-series or geometric-series.

image

「Ratio Test」

Take limit of two terms ratio.

image

「Root Test」

Take the limit of nth root of nth term.

image

「Alternating Series Test」

Test if terms are decreasing, and take limit of nth term.

image

「Absolute Convergence」 & 「Conditional Convergence」

image

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solomonxie commented Jun 23, 2018

nth Term Test

It's also called the nth term divergence test.
The test can only tell if the series is divergent or not. It CAN NOT tell if it converges.

Jump over to Khan academy practice: nth term test
Refer to Khan academy: nth term divergence test

▼Here is the divergent test, very simple:
image

Example

image
Solve:

  • Take the limit of the nth term we get that the limit = 0.
  • According to divergent test rules, we can't conclude anything about it.

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solomonxie commented Jun 23, 2018

Integral Test

►Jump over to have practice at Khan academy: Integral test.
Refer to article from tkiryl: The Integral Test
Refer to Khan academy: Integral Test

▼Refer to awesome article from xaktly: Integral Test

image

image

「Conditions」 of Integral test

Assume the series a𝖓 can be represented as a function f(x).
There are a few limitations for it to use the Integral test:

  • f(x) MUST BE continuous.
  • f(x) > 0. It MUST BE a positive function.
  • f'(x) < 0. It's MUST BE decreasing.

「Using」 Integral test

image

Example

image
Solve:
image

「Understanding」 Integral test

The Integral test has introduced the idea of calculating the total area under the function:

  • The series has step of 1, which means Δx = 1
  • We can sum the areas (which equals the series itself):
    image
  • But when we are to INTEGRATE the function area under the function:
    image
  • The dx is infinitely small rather than a fixed number Δx = 1.
  • As result, the INTEGRAL is almost always greater than the SERIES AREAS.

As been said above, we got this conclusion:
image

Notice: DO NOT use the Integral Test to EVALUATE series, because in general they are NOT equal.

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solomonxie commented Jun 23, 2018

p-series Test

For the the series in form of 1/nᴾ,
the easiest way to determine its convergence is using the p-series test:

image

▼Refer to xaktly: p-series test/harmonic series
image

「Harmonic Series」

Harmonic Series is ∑ 1/n, and Harmonic Series DIVERGES. That's all you need to know.

image

▼Refer to Khan academy: harmonic series diverges
image

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solomonxie commented Jun 23, 2018

❖ Comparison Test

You can understand Comparison Test intuitively as a Sandwich Test.

THIS TEST IS GOOD FOR RATIONAL EXPRESSIONS.

「Direct Comparison Test」

Assume that we have a series a_n, and we're to make up a similar series to it as b_n:

The logic is:

  • If b > a & b converges, then a converges as well.
  • If a > b & b diverges, then a diverges as well.

It's so much easier if you think it graphically.

▼Refer to video: Comparison Test (KristaKingMath)
image
image

▼Refer to xaktly: Comparison Test
image

Example

image
Solve:

  • Assume the asked series as a_n, and we make up a very similar series bigger than it, as b_n:
    image
  • Apply the p-series test we get that the b_n converges.
  • Since a_n < b_n, so according to the Sandwich test (Direct comparison test), a_n converges as well.

「Limit Comparison Test」

Limit comparison test is like an extension when the Direct comparison test won't work.
etc., when we compare a with b, although b converges but a > b, so we can't make any conclusion.
And that's where the limit comparison test comes in place.

The logic is:

  • Take the limit of the division a/b.
  • If the Limit > 0, then they both converges or both diverges.
  • If the Limit ≤ 0, then there's no conclusion.

Jump over to Khan academy for practice: Limit comparison test

▼Refer to video: Limit Comparison Test (KristaKingMath)
image

▼Refer to xaktly: Limit Comparison Test
image

Example

image
Solve:
image

Example

image
Solve:

  • We can't easily tell in the comparison who's greater, so we decide to apply the limit comparison test:
    image

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solomonxie commented Jun 23, 2018

Ratio Test

THIS TEST IS GOOD FOR FACTORIALS.

▼Refer to xaktly: Ratio test / root test

image

Example

image
Solve:

  • Apply the ratio test:
    image
  • L = 0, so the series absolutely converges.

Example

image
Solve:

  • Apply the ratio test, let a_(n+1) / a_n.
  • The trick is to know: (n+1)! = (n+1)·n!, and (n+8)! = (n+8)·(n+7)!
  • Take the limit to get:
    image
  • According to the ratio test, if L = 1, then there's no conclusion.

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solomonxie commented Jun 25, 2018

Root Test

"The root test is used in situations where a series term or part of it is raised to the power of the index variable. "

Notice: The root test isn't a good choice if a series contains factorial terms.
If the root test isn't fairly easy to use, you probably shouldn't use it.
But when it works, it often cuts to convergence or divergence quickly.

▼Refer to xaktly: Ratio test / root test

image

Example

image
Solve:

  • Apply the root test:
    image
  • Since L < 1, so it converges.

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solomonxie commented Jun 26, 2018

❖ Alternating Series Test

It's the test for Alternating series.

►Refer to Khan academy: Alternating series test
►Refer to xaktly: Alternating Series

Alternating Series

It means,
Terms of the series "alternate" between positive and negative.

etc., The alternating harmonic series:
image

The Alternating Series Test

image

The very good example of this test is the Alternating Harmonic Series:

image

▲ It does CONVERGES. (But the Harmonic Series does NOT converge)

Strategy:

  • Take AWAY the Alternating sign (-1)ⁿ:
    image
  • Determine if the rest part is a decreasing series:
    image
  • Take limit of the rest part:
    image
  • If Limit = 0, then the series CONVERGES.
  • If Limit ≠ 0, then the series DIVERGES.

Example

image
Solve:

  • Notice this is an alternating series, so we're to apply the alternating series test.
  • Take away the alternating term, and left with (2/p)ⁿ.
  • So the series only converges if (2/p)ⁿ is decreasing and its limit is 0.
  • And the only way to make it decreasing is to make sure (2/p) < 1.
  • Based on that p value, the limit of (2/p)ⁿ is surely a 0.
  • Therefore, p > 2 makes the series converges.

Example

image
Solve:

  • Notice this is an alternating series, so we're to apply the alternating series test.
  • Take away the alternating term, and left with (2n)ᴾ.
  • So the series only converges if (2n)ᴾ is decreasing and its limit is 0.
  • And the only way to make it decreasing is to make sure p < 0.
  • Based on that p value, the limit of (2n)ᴾ is surely a 0.
  • Therefore, p < 0 makes the series converges.

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solomonxie commented Jun 26, 2018

❖ 「Absolute」 vs. 「Conditional Convergence」

This section is not about calculation, but rather about the logic.

▶Refer to Khan academy: Conditional & absolute convergence

▶ Back to previous note on: p-series test
▶ Practice at Khan academy: determine-absolute-or-conditional-convergence

image

▼We can have a series in Given form and Absolute form:
image

We call the series:

  • Absolute Convergent: if the series converges in BOTH Given form & Absolute form.
  • Conditional Convergent: if the series converges ONLY in Given form BUT NOT in Absolute Form.

etc.,
image

▼Refer to xaktly
image

Example

image
Solve:
image

Example

image
Solve:
image

Example

image
Solve:
image

Example

image
Solve:
image

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solomonxie commented Jun 26, 2018

❖ Error Estimation of Alternating Series

It's also called the Remainder Estimation of Alternating Series.

This is to calculating (approximating) an Infinite Alternating Series:
image

►Jump over to Khan academy for practice: Alternating series remainder

►Refer to The Organic Chemistry Tutor: Alternate Series Estimation Theorem
►Refer to Mathonline: Error Estimation for Approximating Alternating Series
►Refer to mathwords: Alternating Series Remainder

The logic is:

  • First to test the series' convergence.
  • If the series CONVERGES, then we can proceed to calculate it by Error Estimation Theorem. Otherwise we aren't able to.
  • We can express the series as the sum of partial sums & infinite remainder:
    image
    (▲ Sn is the first n terms, and Rn is from the n+1 term to the rest terms.)
  • And the "structure" in the partial sum & remainder is:
    image
  • With a little twist, we will get the whole idea:
    image
    (▲Since the Rn is the gap between S & Sn, so we call it The Error)
  • ▼ And the theorem is: The Remainder MUST NOT be greater than its first term:
    image

▼Actual sum = Partial sum + Remainder: refer to Khan academy: Alternating series remainder

image

「Sign」 & 「Size」 of Error

►Refer to Khan academy: Alternating series remainder
►Refer to Khan academy: Worked example: alternating series remainder

For the Remainder series, its FIRST TERM is always DOMINATING the whole remainder:

  • It dominates the remainder's SIGN: positive or negative.
  • It dominates the remainder's SIZE: the whole remainder's absolute value CAN'T BE greater than the first term.

Based on the error's sign,
we could tell the approximated series is UNDERESTIMATED or OVERESTIMATED:

  • If Error > 0, then the approximated series is Underestimate.
  • If Error < 0, then the approximated series is Overestimate.

Bound the Error (accuracy control)

The error bound regards to the accuracy of the approximated series, and we want to control the accuracy before approximation.

►Refer to Khan academy: Worked example: alternating series remainder

We have 2 ways to bound the error in a range:

  • Set up how small we want the error to be, or
  • Set up how many terms we want to have in the partial sum Sn.

Bound by terms

The Larger n → The smaller gap → The lesser Error → The more accurate.

Strategy:

  • We could set the partial sum to include a certain number of terms, etc. 100 terms
  • And the first term of Remainder should be the 101st term.
  • The error bound, or the error is dominated by the first term.
  • So we say the error bound IS the value of the 101st term.

Bound the error

To bound the error in a range, we often say:

  • "Approximate the series to the 2 decimal places",
  • "Let the error be less than 0.01",
  • "We want the accuracy within ±0.01"

What they mean are the same:
image

▲ And by solving the inequality,
we will get the scope for n,
then get the Smallest Integer of n in that scope.

Example

image
Solve:

  • First to notice, the partial sum is already set to 100 terms, so we're to control accuracy by bound the terms.
  • So the error should be from the 101st term to infinity.
  • But the error bound is actually dominated by the first term of the error.
  • So the error bound = the value of 101st term:
    image
  • We could say that: The error bound is negative, and negative error causes overestimation.

Example

image
Solve:

  • It's clear this is a alternating series.
  • So we want to do the alternating series test first, and it passed, which means it converges.
  • Since the series converges, we can do further approximation.
  • See that we don't know how many terms are in the partial sum, and only know how much accurate we'd like.
  • So we're to approximate by bound the error, and find out the terms.
  • Apply the Error Approximation Theorem, assume the first term of remainder is a_(n+1):
    image
  • Solve out the inequality to get n ≥ 999,999
  • And 999,999 the smallest integer of n to make the series converges with 2 decimal accuracy.

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solomonxie commented Jun 27, 2018

❖ Error Estimation Theorem

The Error Estimation Theorem is not only for alternating series, but available for all infinite series.

▼Boundary of estimating series: refer to Khan academy: Series estimation with integrals

image

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solomonxie commented Jun 29, 2018

❖ Interval of Convergence

When we see the series as a function, we can actually specify an interval for the function so that the series certainly converges over this interval.

►Jump over to have practice at Khan academy: Interval of convergence

The method is kind of like finding the interval of an ordinary function:

  • The term of series can't be 0, so set a_n ≠ 0 and solve it to get the condition.
  • Take some convergence tests, etc. ratio test.
  • Calculate to get the condition that makes the test passes.
  • Substitute the endpoints of the interval back in the function and see if it also converges.

Example

image
Solve:

  • The term can't be zero, so:
    image
  • And further more, take a ratio test which makes it converges:
    image
  • Calculate the ratio test to get the interval:
    image
  • Get the interval for x:
    image
  • Test the endpoints for this interval:
    image
  • In conclusion, the interval of convergence is:
    image

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solomonxie commented Jun 29, 2018

❖ Power Series

Try to think Power series = Geometric Series.

►Refer to Math24: Power series

Power series is actually the Geometric series in a more general and abstract form.

image

For easier to remember it, that could be simplified as:

image

In this function it's critical to know that:
a_n IS A CONSTANT NUMBER! NOT A VARIABLE !

Differentiate 「Power series」

►Refer to Khan academy: Differentiating power series

We have 2 ways to differentiate series, they work same way:

  • One way is to expand the series with real numbers (1,2,3...) and take their derivatives:
    image
    (▲Note that this is constantly true for power series.)
  • Another way is to calculate the term's derivative and then plug in the real numbers (1,2,3):
    image

Either way will do, it depends on the actual equation for you to choose which way you're gonna use.

Example

image
Solve:

  • Let's organize this function to make it clear:
    image

Example

image
Solve:

  • Notice that: If we plug in the x=0 at beginning, everything will be 0 and we don't have anything to calculate.
  • So Let's keep the x in the terms until the last step.
  • It's easier to expand the series with real numbers, and we're to try 3 or 4 terms in this case.
  • Because at the end of it you'll notice, if we're doing Third Derivative, then more than 3 terms will just bring more 0s.
  • First we're to organize the function in the standard power series form:
    image
  • And the constant number a_n in the function is:
    image
  • Let's plug in the real number (0,1,2,3,4) for n to expand the series:
    image
  • Based on that we can take the derivatives:
    image
  • Now we've got the third derivative, so let's plug in the x=0 and see what we get:
    image
  • And that's the answer.
  • And now you know why in this case it's a waste to calculate more terms.

Integrate 「Power Series」

Example

image
Solve:

  • Knowing that Integrate a series can be turned to a series of Integrations of terms:
    image
  • Integrate the terms:
    image
  • And we get a simple Geometric series. So that we can apply the formula of calculating geometric series:
    image
  • Let's apply the formula:
    image

「Integrals & derivatives」 of functions with 「known power series」

► Jump over to have practice at Khan academy.

Example

image
Solve:

  • Let's assume the function of the series above is f(x), and the series below is g(x)
  • It's clear to see that the g(x) is the Antiderivativeof thef(x)`.
  • So we just need to integrate the function of the f(x):
    image
  • We see that the antiderivative can represent the g(x), but only with the C in it:
    image
  • So we need to solve for C. The easiest way is to plug in 0 for g(x):
    image
  • Then the answer is:
    image

Example

image
Solve:

  • Set the two series as f(x) and g(x).
  • It's clear that g(x) = -f'(x).
  • So by differentiate f(x) we will get:
    image
  • Then -f'(x) would be:
    image

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solomonxie commented Jul 2, 2018

❖ Taylor Series

Taylor series, or Taylor polynomial is a series that can REPRESENT a function,
regardless what function it is.

▼Refer to 3Blue1Brown for animation & intuition: Taylor series | Chapter 10, Essence of calculus

"Taylor Series is one of the most powerful tools Math has to offer for approximating functions." - 3Blue1Brown

►Refer to Khan academy: Taylor & Maclaurin polynomials intro (part 1)
▼Refer to xaktly: Taylor Series
image

image
(▲ C represents the centre where we're centred at to approximate the function.)

▲ Notice: The Taylor Series is a Power Series, which means we can use a lot of techniques of power series on this to operate it easily.

We could expand it and make it clearer ▼:
image

The main purpose of using a Taylor Polynomial is to REPLACE the original function with a polynomial, which it is easy to work with.

etc., we can express the function f(x) = eˣ as ▼:
image

More importantly, by adding more & more terms into the polynomial, we can approximate the function more precisely:

►Refer to joseferrer: Mathematical explanation - Taylor series
►For More animation, visit Desmos: Taylor Series Visualization

97f5384c9b8d6ceebf3b894efc106adb

Example

image
Solve:

  • First to know the formula of Taylor Series centred at x=1:
    image
  • The problem is asking the coefficient of (x-1)³, means all the rest part in the formula, which is:
    image
  • And it also means the n=3, so the coefficient becomes:
    image
  • Let's evaluate the f'''(1):
    image
  • So the coefficient is:
    image

Example

image
Solve:

  • Let's express the Taylor polynomial to the nth degree as:
    image
  • Since it's asking for the series to the 3rd degree, then it becomes:
    image
  • And we only need to find out every degree of derivatives, and we will get:
    image
  • So the Taylor polynomial then is:
    image

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solomonxie commented Jul 2, 2018

❖ Maclaurin Series

It's also called Maclaurin polynomials.

Maclaurin series is a special case of Taylor Series which centres at x=0.

image

▼Expand it we'll understand it better:
image

▼Here is a graph we're trying to approximate a function centred at x=0:
image

Example

image
Solve:

  • This problem is to test if you're familiar with the Maclaurin Series Formula.
  • Let's ignore all others and only see the asked x⁴.
  • x⁴ means we're gonna find out the term of the 4th derivative, and plug in 4 into the formula, we'll get the term:
    image
  • It's asking for the coefficient of x⁴, which is the rest part in that formula for the term:
    image
  • And we only need to find out what is the value of the 4th derivative.
  • By the given formula of gᴺ(0), we can get:
    image
  • So the coefficient will be:
    image

Example

image
Solve:

  • We know the Maclaurin series is a Taylor series centred at x=0, and the formula is:
    image
  • It's told to list 4 terms, so we plug in the given value of f', f'', f''' and get:
    image
  • And we get the answer:
    image

Evaluate 「Maclaurin Series」

To evaluate a Maclaurin series, we need to convert the series to a function, and then evaluate the function.

▼Jump forward to have a look at the note: Maclaurin Series of Common functions

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solomonxie commented Jul 3, 2018

❖ Lagrange Error Bound

It's also called the Lagrange Error Theorem, or Taylor's Remainder Theorem.

To approximate a function more precisely, we'd like to express the function as a sum of a Taylor Polynomial & a Remainder.

image
(▲ For T is the Taylor polynomial with n terms, and R is the Remainder after n terms.)

▲Jump back to review the note on Error estimation Theorem.

►Jump over to have practice at Khan academy: Lagrange Error Bound.

The tricky part of that expression is to "preset" the accuracy of the Error, aka. the Remainder.

For bounding the Error, out strategy is to apply the Lagrange Error Bound theorem.

Simply saying, the theorem is:

  • If a function's ALL DERIVATIVES are bounded by a number over the interval (C, x):
    image
    (▲ for C is the centre of approximation)
  • where the max value of all derivatives of the function is:
    image
    (▲ for z is any value between C and x makes the derivative to the max)
    (▲ and note that: the input has to be n+1 )
  • then the function's Remainder MUST satisfy this theorem:
    image

Example

image
Solve:

  • This problem is to approximate a function with Taylor Polynomial.
  • To do so, we're to set use the Error estimation method, which first to set up the equation:
    image
  • In this case, it is:
    image
  • Since it's only asking for the error bound, so we only focus on the Error Rn.
  • We want to apply the Lagrange Error Bound Theorem, and bound it to 0.001:
    image
  • For those unknowns variables in the theorem, we know that:
    • The approximation is centred at 1.5π, so C = 1.5π.
    • The input of function is 1.3π, so x = 1.3π.
    • For The M value, because all the derivatives of the function cos(x), are bounded to 1 even without an interval , so let's say the max value M = 1.
  • Therefore, the formula of this theorem becomes:
    image
  • Unfortunately, at this moment we don't have easier method to solve for n except trying some numbers in:
    image
  • We could see that, with the degree gets larger and larger, the Error becomes smaller and smaller.
  • Only until n=4, which means the 4th derivative, the Error is less than 0.001.
  • So the answer is 4th derivative.

Example

image
Solve:

  • Same with the problem above, we want to apply the Lagrange Error Bound Theorem, and bound it to 0.001:
    image
  • For those unknowns variables in the theorem, we know that:
    • The approximation is centred at 0 because it's told as a Maclaurin Series , so C = 0.
    • The input of function is -0.95, so x = -0.95.
    • The interval is (C, x) or (x, C), which is (-0.95, 0) in this case.
  • For The M value, since all the derivatives of is just , and is unbounded at all, so we're to examine the Max value over the interval (-0.95, 0)
  • With the help from Desmos Calculator, we know that over the interval (-0.95, 0), the max value of is e⁰ = 1:
    image
  • So boundary is M = 1.
  • Therefore, the formula of this theorem becomes:
    image
  • In this case, we need to try some numbers for n to get the desired value:
    image
  • After tried n=5 and n=6, we could see that only until n=6, which means the 6th derivative, the Error is less than 0.001.
  • So the answer is 6th derivative.

Example

image
Solve:

  • Same with the problem above, we want to apply the Lagrange Error Bound Theorem, and bound it to 0.001:
    image
  • For those unknowns variables in the theorem, we know that:
    • The approximation is centred at 2, so C = 2.
    • The input of function is 2.5, so x = 2.5.
    • The interval then is (2, 2.5).
  • For the M value, it's not easy to figure out, but we've been told the formula for derivative.
    image
  • So the expression for M would be:
  • Let's directly plug in the M expression into the Remainder:
    image
  • With the help from Desmos grapher we know that when within the interval 2≤ z ≤ 2.5, that z=2 makes the formula to the max:
    image
  • So let's set the inequality:
    image
  • After trying out some number for n, we get that n ≥ 3.

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solomonxie commented Jul 6, 2018

Finding Taylor series for a function [DRAFT]

►Refer to Khan academy's unit: Finding Taylor or Maclaurin series for a function

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solomonxie commented Jul 6, 2018

Function as a 「Geometric Series」

Example

image
Solve:

  • This function looks very alike the Infinite Geometric function:
    image
  • So we extract the key numbers out:
    • First term: a = 1
    • Common Ratio: r = -x²
  • So the infinite geometric series would be:
    image
  • Expand the formula to be:
    image

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solomonxie commented Jul 6, 2018

❖ Maclaurin Series of Common functions

►Refer to Wiki: List of Maclaurin series of some common functions

Maclaurin series of these common functions are very useful, which we really want to memorize.

Function Maclaurin Series
sin(x) image
cos(x) image
tan(x) image
sec(x) image
image
Geometric series 1 image
Geometric series 2 image
Geometric series 3 image

Example

image
Solve:

  • This Maclaurin series has a clear pattern of cos(x) Maclaurin series:
    image
  • In this case, it's:
    image
  • So we could convert the series back to the trig function:
    image

Example

image
Solve:
image

Example

image
Solve:

  • Looks quite a complicated series, but we look at the factorial dominators, we found that resembles the pattern of the sin(x)'s maclaurin series.
  • By comparing to the sin(x) maclaurin series, we notice that x = π/2, and all the rest are the same.
  • So rewrite the series to the function:
    image
  • And we get the result is 1.

Example

image
Solve:

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solomonxie commented Jul 6, 2018

「Euler's Formula」 & 「Euler's Identity」

"The Euler's identity connects all of these FUNDAMENTAL NUMBERS in some mystical way that shows that there's some connectedness to the UNIVERSE.
If this does not blow your mind, you have no emotion." - Sal Khan

▶︎Refer to the most well-known lecture from Sal Khan: Euler's formula & Euler's identity

Euler's formula:
image

image

Euler's identity:
image

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solomonxie commented Jul 30, 2018

Multivariable functions [DRAFT]

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solomonxie commented Jul 30, 2018

3D Vector Fileds [DRAFT]

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solomonxie commented Jul 30, 2018

Parametric Functions [DRAFT]

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solomonxie commented Jul 30, 2018

Partial derivatives [DRAFT]

"...this concept is nothing to be intimidated by. Partial differentiation is essentially just taking a multi dimensional problem and pretending that it's just a standard 1D problem when we consider each variable separately."

Refer to Coursera: Variables, constants & context

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solomonxie commented Jul 30, 2018

Gradient [DRAFT]

Now we take many partial derivatives out from a function, and we need to pack them together in a form.
We call the form of pack The gradient of the function.

image

For example:
image

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solomonxie commented Jul 30, 2018

Total derivatives [DRAFT]

Refer to Coursera: Differentiate with respect to anything

image

image

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solomonxie commented Jul 30, 2018

Jacobian [DRAFT]

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solomonxie commented Aug 1, 2018

Directional derivative [DRAFT]

It is the more general form of the Partial derivative, because it not only give the derivatives with respects to axes, but to any direction.

Refer to Khan academy: Directional derivative
Refer to youtube: Directional Derivatives and The Gradient

The Partial derivatives only give the SLOPE in x direction and y direction and more directions with respect to the variables, but what if we want to have a SLOPE in ANY DIRECTION?

So the Directional derivative comes in place to achieve that.

How to do this:

  • Pick a direction: We pick the direction θ
  • Represent the direction by a unit vector in form of u = cosθ·i + sinθ·j, because cos²θ + sin²θ = 1.
  • Apply the directional derivative's formula:
    image
    (Dᵤ means Directional Derivative of a Unit Vector u, f𝓍 means Partial derivative respects to x, and f𝑦 means Partial derivative respects to y.)

image

The Directional derivative is the Dot product of The Gradient & Unit vector:

image

Which comes from:

image

image

image

image
image

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